Hindi, asked by srikar4312, 1 year ago

मान ज्ञात कीजिए : (i) $(3^0 + 4^{-1}) \times 2^2$ (ii) $(2^{-1} \times 4^{-1}) \div 2^{-2}$ (iii) $(\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} +(\frac{1}{4})^{-2}$ (iv) $(3^{-1} + 4^{-1} + 5^{-1})^0$ (v) $\{(\frac {-2}{3})^{-2}\}^2$

Answers

Answered by nikitasingh79

Answer with Explanation:

(i) (3⁰ + 4⁻¹) * 2²

= (1 + 1/4) ₓ 2²

[a⁰ = 1 and a^⁻m = 1/ a^m]

= (5/4) ₓ 2²

= (5/4) ₓ 4

= 5 ₓ 1

= 5

(ii) (2⁻¹ ₓ 4⁻¹) ÷ 2⁻²

= (1/2 ₓ 1/4) ÷ 2⁻²

[a^⁻m = 1/ a^m]

= (1/8) ÷ 2⁻²

= (1/2³) ÷ 2⁻²

= 2⁻³ ÷ 2⁻²

= 2⁻³ ⁻⁽ ⁻² ⁾

[am ÷ an = am-n]

= 2⁻³ ⁺²

= 2⁻¹

= 1/2

[a^⁻m = 1/ a^m]

(iii) (1/2)⁻² + (1/3)⁻² + (1/4)⁻²

= (2/1)² + (3/1)² + (4/1)²

[a^⁻m = 1/ a^m]

= 2² + 3² + 4²

= 4 + 9 + 16

= 29

(iv) (3⁻¹ + 4⁻¹ + 5⁻¹)⁰

= 1

[(a + b)⁰ = 1]

(v) {(-2/3)⁻²}²

= {(-3/2)²}²

[a^⁻m = 1/ a^m]

= (-3/2)²ˣ²

= (-3/2)⁴

= -3⁴/2⁴

= 81/16

आशा है कि यह उत्तर आपकी अवश्य मदद करेगा।।।।

इस पाठ से संबंधित कुछ और प्रश्न :

सरल कीजिए और उत्तर को धनात्मक घातांक के रूप में व्यक्त कीजिए। (i) $(- 4)^5 \div (- 4)^8$ (ii) $(\frac {1}{2^3})^2$ (iii) $(-3)^4 \times (\frac {5}{3})^4$ (iv) $(3^{- 7} \div 3^{-10}) \times 3^{- 5}$ (v) [tex] 2^{- 3} \times (-7)^{- 3} [/tex